A car moving $18.2 m / s$ starts to coast UP a frictionless ${10.0}^{\circ}$ hill. How far does it roll before coming to a stop?
(Unit = m)

Answer

Expert–verified

Hide Steps

Answer

4. Solve for the distance up the hill (d): $d = \frac{h}{\sin ({10.0}^{\circ})}$

Steps

Step 1 ：1. Calculate the initial gravitational potential energy (PE): $P E_{i} = m g h$

Step 2 ：2. Calculate the final gravitational potential energy (PE): $P E_{f} = P E_{i} - \frac{1}{2} m v_{i}^{2}$

Step 3 ：3. Solve for the height (h): $h = \frac{P E_{f}}{m g}$

Step 4 ：4. Solve for the distance up the hill (d): $d = \frac{h}{\sin ({10.0}^{\circ})}$